class Solution {
    int dx[4] = {0, 0, 1, -1};
    int dy[4] = {1, -1, 0, 0};
    int m, n;
public:
    void solve(vector<vector<char>>& board) {
        m = board.size(), n = board[0].size();

        // 1. 先处理边界上的 'O' 连通块，全部修改成 '.'
        // 处理第一行和最后一行
        for(int j = 0; j < n; j++)
        {
            if(board[0][j] == 'O')
                bfs(board, 0, j);
            if(board[m - 1][j] == 'O')
                bfs(board, m - 1, j);
        }
        // 处理第一列和最后一列
        for(int i = 0; i < m; i++)
        {
            if(board[i][0] == 'O')
                bfs(board, i, 0);
            if(board[i][n - 1] == 'O')
                bfs(board, i, n - 1);
        }

        // 2. 还原
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
                if(board[i][j] == 'O')
                    board[i][j] = 'X';
                else if(board[i][j] == '.') // 将 '.' 还原成 'O'
                    board[i][j] = 'O';
    }

    void bfs(vector<vector<char>>& board, int i, int j)
    {
        queue<pair<int, int>> q;
        q.push({i, j});
        board[i][j] = '.';

        while(!q.empty())
        {
            auto [a, b] = q.front();
            q.pop();
            for(int k = 0; k < 4; k++)
            {
                int x = a + dx[k], y = b + dy[k];
                if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O')
                {
                    q.push({x, y});
                    board[x][y] = '.';
                }
            }
        }
    }
};